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Unformatted text preview: Portillo, Tom Homework 9 Due: Nov 1 2006, 3:00 am Inst: David Benzvi 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Evaluate the double integral I = Z 3 2 Z 1 x y 2 dx dy. Correct answer: 3 . 16667 . Explanation: The double integral I can be written as the repeated integral I = Z 3 2 Z 1 x y 2 dx dy, integrating first with respect to x . Now after use of the Fundamental theorem of integral calculus, the inner integral becomes h x 2 y 2 2 i 1 = 1 2 y 2 . Thus I = Z 3 2 1 2 y 2 dy = h 1 2 y 3 3 i 3 2 . Consequently, I = 19 6 . keywords: 002 (part 1 of 1) 10 points Evaluate the integral I = Z 2 1 Z 4 1 ( x 2 2 xy ) dydx. 1. I = 21 2 2. I = 25 2 3. I = 13 2 4. I = 27 2 correct Explanation: The integral can be written in iterated form I = Z 2 1 Z 4 1 ( x 2 2 xy ) dy dx. Now Z 4 1 ( x 2 2 xy ) dy = h x 2 y xy 2 i 4 1 = (4 x 2 16 x ) ( x 2 x ) = 3 x 2 15 x. But then I = Z 2 1 (3 x 2 15 x ) dx = h x 3 15 x 2 2 i 2 1 . Consequently, I = 27 2 . keywords: definite integral, iterated integral, polynomial function, 003 (part 1 of 1) 10 points Evaluate the iterated integral I = Z 2 1 n Z 2 1 x y + y x dy o dx. 1. I = 2ln 3 2 2. I = 3ln2 correct 3. I = 3ln 3 2 4. I = 3 2 ln3 5. I = 3 2 ln2 Portillo, Tom Homework 9 Due: Nov 1 2006, 3:00 am Inst: David Benzvi 2 6. I = 2ln3 Explanation: Integrating with respect to y keeping x fixed, we see that Z 2 1 x y + y x dy = x ln y + y 2 2 x 2 1 = (ln2) x + 3 2 1 x . Thus I = Z 2 1 (ln2) x + 3 2 1 x dx = x 2 2 ln2 + 3 2 ln x 2 1 . Consequently, I = 3ln2 . keywords: 004 (part 1 of 1) 10 points Evaluate the double integral I = Z 3 2 Z 2 e x y dxdy . 1. I = e 3 e 2 e 1 1 2. I = e 3 e 2 + e 1 + 1 3. I = e 3 e 2 e 1 + 1 correct 4. I = e 3 + e 2 e 1 + 1 Explanation: After integration with respect to x , I = Z 3 2 e x y / 2 dy = Z 3 2 ( e 2 y e y ) dy . But then, after integrating next with respect to y we see that I =  e 2 y + e y / 3 2 = e 1 + e 3 ( 1 + e 2 ) . Consequently, I = e 3 e 2 e 1 + 1 . keywords: 005 (part 1 of 1) 10 points Determine the value of the double integral I = Z Z A 3 xy 2 4 + x 2 dA over the rectangle A = n ( x,y ) : 0 x 3 , 1 y 1 o , integrating first with respect to y . 1. I = ln 13 8 2. I = ln 4 13 3. I = 1 2 ln 13 8 4. I = 1 2 ln 13 4 5. I = 1 2 ln 4 13 6. I = ln 13 4 correct Explanation: The double integral over the rectangle A can be represented as the iterated integral I = Z 3 Z 1 1 3 xy 2 4 + x 2 dy dx, Portillo, Tom Homework 9 Due: Nov 1 2006, 3:00 am Inst: David Benzvi 3 integrating first with respect to y . Now after integration with respect to y with x fixed, we see that Z 1 1 3 xy 2 4 + x 2 dy = h xy 3 4 +...
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This note was uploaded on 02/18/2010 for the course MATH 408L taught by Professor Benzvi during the Spring '06 term at North Texas.
 Spring '06
 Benzvi

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